Question

The sun's energy that actually reaches the earth is about how many kilowatt hours per day

Answer 1

Hey Mate Here Is Your Answer

Note that we measure energy in units of Watt-hours. A watt is not a unit of energy, it is a measure of power.

ENERGY = POWER x TIME

1 Kilowatt Hour = 1KWH = 1000 watts used in one hour = 10 100 watt light bulbs left on for an hour

Incident Solar Energy on the ground:

 Average over the entire earth = 164 Watts per square meter over a 24 hour day  So the entire planet receives 84 Terrawatts of Power  our current worldwide consumption is about 12 Terrawatts  so is this a solution?



and remember, little of the world current runs on 
renewable energy sources



There is a large amount of infrastructure (e.g. cost) required to convert frompotential to deliverable energy.

8 hour summer day, 40 degree latitude  600 Watts per sq. meter

So over this 8 hour day one receives:

8 hours x 600 watts per sq. m = 4800 watt-hours per sq. m which equals 4.8 kilowatt hours per sq. m

This is equivalent to 0.13 gallons of gasoline

For 1000 square feet of horizontal area (typical roof area) this is equivalent to 12 gallons of gas or about 450 KWHBut to go from energy received to energy generated requires conversion of solar energy into other forms (heat, electricity) at some reduced level of efficiency.








We will talk more about PV cells in detail later. For now the only point to retain is that they are quite low in efficieny!

Collection of Solar Energy

Amount of captured solar energy depends critically on orientation of collector with respect to the angle of the Sun.

Under optimum conditions, one can achieve fluxes as high as 2000 Watts per sq. meter

In the Winter, for a location at 40 degrees latitude, the sun is lower in the sky and the average flux received is about 300 Watts per sq. meter

A typical household Winter energy use is around 2000-3000 KWHs per month or roughly 70-100 KWH per day.

Assume our roof top area is 100 square meters (about 1100 square feet).

In the winter on a sunny day at this latitude (40o) the roof will receive about 6 hours of illumination.

So energy generated over this 6 hour period is:

300 watts per square meter x 100 square meters x 6 hours

= 180 KWH (per day)  more than you need.

But remember the efficiency problem:

5% efficiency  9 KWH per day10% efficiency  18 KWH per day20% efficiency  36 KWH per day

At best, this represents 1/3 of the typical daily Winter energy usage and it assumes the sun shines on the rooftop for 6 hours that day.

With sensible energy conservation and insulation and south facing windows, its possible to lower your daily use of energy by about a factor of 2. In this case, if solar shingles become 20% efficient, then they can provide 50-75 % of your energy needs

Another example calculation for Solar Energy which shows that relative inefficiency can be compensated for with collecting area.

A site in Eastern Oregon receives 600 watts per square meter of solar radiation in July. Asuume that the solar panels are 10% efficient and that the are illuminated for 8 hours.

How many square meters would be required to generate 5000 KWH of electricity?

 each square meter gives you 600 x.1 = 60 watts

 in 8 hours you would gt 8x60 = 480 watt-hours or about .5 KWH per square meter

 you want 5000 KWH

 you therefore need 5000/.5 = 10,000 square meters of collecting area


Solar Energy: Collection, Energy Generation and Heat Transfer

Rishikesh.